Solve for $x$ and $y$ using substitution. ${3x+2y = -1}$ ${x = 2y+5}$
Since $x$ has already been solved for, substitute $2y+5$ for $x$ in the first equation. ${3}{(2y+5)}{+ 2y = -1}$ Simplify and solve for $y$ $6y+15 + 2y = -1$ $8y+15 = -1$ $8y+15{-15} = -1{-15}$ $8y = -16$ $\dfrac{8y}{{8}} = \dfrac{-16}{{8}}$ ${y = -2}$ Now that you know ${y = -2}$ , plug it back into $\thinspace {x = 2y+5}\thinspace$ to find $x$ ${x = 2}{(-2)}{ + 5}$ $x = -4 + 5$ ${x = 1}$ You can also plug ${y = -2}$ into $\thinspace {3x+2y = -1}\thinspace$ and get the same answer for $x$ : ${3x + 2}{(-2)}{= -1}$ ${x = 1}$